Solve for $r$, $ -\dfrac{1}{5r - 5} = \dfrac{6}{r - 1} + \dfrac{4r - 10}{2r - 2} $
Solution: First we need to find a common denominator for all the expressions. This means finding the least common multiple of $5r - 5$ $r - 1$ and $2r - 2$ The common denominator is $10r - 10$ To get $10r - 10$ in the denominator of the first term, multiply it by $\frac{2}{2}$ $ -\dfrac{1}{5r - 5} \times \dfrac{2}{2} = -\dfrac{2}{10r - 10} $ To get $10r - 10$ in the denominator of the second term, multiply it by $\frac{10}{10}$ $ \dfrac{6}{r - 1} \times \dfrac{10}{10} = \dfrac{60}{10r - 10} $ To get $10r - 10$ in the denominator of the third term, multiply it by $\frac{5}{5}$ $ \dfrac{4r - 10}{2r - 2} \times \dfrac{5}{5} = \dfrac{20r - 50}{10r - 10} $ This give us: $ -\dfrac{2}{10r - 10} = \dfrac{60}{10r - 10} + \dfrac{20r - 50}{10r - 10} $ If we multiply both sides of the equation by $10r - 10$ , we get: $ -2 = 60 + 20r - 50$ $ -2 = 20r + 10$ $ -12 = 20r $ $ r = -\dfrac{3}{5}$